$h(x)=-\dfrac{2}{x+3}$. On which intervals is the graph of $h$ concave down? Choose 1 answer: Choose 1 answer: (Choice A) A $(-\infty,-3)$ only (Choice B) B $(-\infty,3)$ only (Choice C) C $\left(\dfrac23,\infty\right)$ only (Choice D) D $(-3,\infty)$ only
Explanation: We can analyze the intervals where $h$ is concave up/down by looking for the intervals where its second derivative $h''$ is positive/negative. This analysis is very similar to finding increasing/decreasing intervals, only instead of analyzing $h'$, we are analyzing $h''$. The second derivative of $h$ is $h''(x)=-\dfrac{4}{(x+3)^3}$. $h''$ is never equal to $0$. $h''$ is undefined for $x=-3$. Therefore, our only point of interest is $x=-3$. Our point of interest divides the domain of $h$ (which is all numbers except for $-3$ ) into two intervals: $\llap{-}7$ $\llap{-}6$ $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $(-\infty, \llap{-}3)$ $( \llap{-}3,\infty)$ Let's evaluate $h''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h''(x)$ Verdict $(-\infty,-3)$ $x=-4$ $h''(-4)=4>0$ $h$ is concave up $\cup$ $(-3,\infty)$ $x=-2$ $h''(-2)=-4<0$ $h$ is concave down $\cap$ In conclusion, the graph of $h$ is concave down over the interval $(-3,\infty)$ only.